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Here $D=I$, so the Jacobi iteration matrix isThe all-ones vector is an eigenvector with eigenvalue $-2\alpha$. On its two-dimensional orthogonal complement, the coordinates sum to zero and $Hv=\alpha v$, so the other eigenvalue is $\alpha$ with multiplicity two. ThereforeBy Jacobi convergence for a three-by-three equicorrelation matrix, convergence occurs exactly forAt either endpoint the spectral radius is one, so convergence for arbitrary initial data fails.

Away from the rim's $O(h)$ edge region, the geometry and uniform injection have no radial scale other than the factor required by continuity. It is therefore consistent to take $w=w(z)$. Regularity at $r=0$ then lets the continuity equation integrate toThe no-slip boundary condition and prescribed normal velocities areSince $p$ is independent of $z$, the radial equation implies that $w^{\prime \prime \prime }$ is constant. The four boundary conditions giveand henceUsing $w^{\prime \prime \prime }=12V/h^3$ in $p_r=\mu u_{zz}$ givesTaking the pressure at the rim to be atmospheric, $p(R)=p_0$, produces the porous-plate lubrication cushion pressureThe upward pressure force balances the disc's weight:Since $\epsilon =h/R$,

Expanding the two covariant derivatives of a vector, the second partial derivatives cancel in the commutator. The remaining derivatives and products of Christoffel symbols combine into the coordinate definition of the Riemann curvature tensor, giving the Ricci identityFor a type-$(2,0)$ tensor, the connection acts on both upper indices, so the curvature commutator on a contravariant tensor is

Keep $\mathbf{A}=(0,Bx,0)$ and choose the scalar potential $\varphi =-Ex$. ThenFor $p_z=0$, completing the square givesHence the spectrum on ${\mathbb{R}}^3$ is

For a translationally invariant dispersion, the group velocity is $v_y={\hslash }^{-1}\partial E_n/\partial k$. Thereforein agreement with the electric-cross-magnetic-field drift $\mathbf{E}\times \mathbf{B}/B^2$.

In the rectangle, $k=2\pi r/L_y$ as before. The energy now changes with $r$, so the guiding-centre degeneracy is lifted, while the number of states in each tilted band remains approximately $|qB|A/(2\pi \hslash )$. Adjacent states within a formerly degenerate level have the electric-field splitting of a Landau level in a rectangleThus this is the ground-to-first-excited gap when the lowest Landau band contains at least two allowed guiding centres and $\Delta E_{\mathrm{electric}}<\hslash {\omega }_c$. Without that implicit weak-field or large-sample condition, the exact full spectral gap is

An odd composite number $N$ with $\gcd (b,N)=1$ is a Fermat pseudoprime to base $b$ whenFor $N=35$, the Chinese remainder theorem for unit groups reduces this condition to congruences modulo $5$ and $7$. Modulo $5$, Fermat's little theorem givesso $b^{34}\equiv 1(\mathrm{mod}5)$ exactly when $b\equiv \pm 1(\mathrm{mod}5)$. Modulo $7$, the unit group has order $6$, soThe equation $b^4=1$ in the cyclic group $(\mathbb{Z}/7\mathbb{Z}{)}^{\times }$ has $\gcd (4,6)=2$ solutions, namely $b\equiv \pm 1(\mathrm{mod}7)$.

Combining the two independent sign choices by the Chinese remainder theorem givesand hence